The slope of the tangent to a curve C: y = y(x) at any point (x, y)

The slope of the tangent to a curve C: y = y(x) at any point (x, y) on it is $\frac{{2{e^{2x}} - 6{e^{ - x}} + 9}}{{2 + 9{e^{ - 2x}}}}$ . If C passes through the points $\left( {0,\frac{1}{2} + \frac{\pi }{{2\sqrt 2 }}} \right)$ and $\left( {\alpha ,\frac{1}{2}{e^{2\alpha }}} \right)$

Then ${e^\alpha }$ is equal to

$\frac{{3 + \sqrt 2 }}{{3 - \sqrt 2 }}$
$\frac{3}{{\sqrt 2 }}\left( {\frac{{3 + \sqrt 2 }}{{3 - \sqrt 2 }}} \right)$
$\frac{1}{{\sqrt 2 }}\left( {\frac{{\sqrt 2 + 1}}{{\sqrt 2 - 1}}} \right)$
$\frac{{\sqrt 2 + 1}}{{\sqrt 2 - 1}}$

Solution:

Tip for solving this question:

Use the formula of the slope of the tangent.

On integration, find the value of y.

Then using points find ${e^\alpha }$

Step 1 of 3:

As we know, slope of tangent = $\frac{{dy}}{{dx}}$

Given, Slope of tangent = $\frac{{2{e^{2x}} - 6{e^{ - x}} + 9}}{{2 + 9{e^{ - 2x}}}}$

$\therefore \frac{{dy}}{{dx}} = \frac{{2{e^{2x}} - 6{e^{ - x}} + 9}}{{2 + 9{e^{ - 2x}}}}$

$dy = {e^{2x}} - \frac{{6{e^{ - x}}}}{{2 + 9{e^{ - 2x}}}}dx$ ……….(1)

Step 2 of 3:

Now, integrate equation (1)

$\int {dy = \int {{e^{2x}} - \frac{{6{e^{ - x}}}}{{2 + 9{e^{ - 2x}}}}dx} }$

$latex\int {dy = \int {{e^{2x}} – \frac{{6{e^{ – x}}}}{{2 + 9{e^{ – 2x}}}}dx} } $

$\Rightarrow y = \int {{e^{2x}}dx} - \int {\frac{{6{e^{ - x}}}}{{2 + 9{e^{ - 2x}}}}dx}$

$\Rightarrow y = \frac{{{e^{2x}}}}{2} - 3\int {\frac{{{e^{ - x}}}}{{1 + {{\left( {\frac{{3{e^{ - x}}}}{{\sqrt 2 }}} \right)}^2}}}dx}$ .

$\begin{gathered} {e^{ - x}} = t \hfill \\ \Rightarrow - {e^{ - x}}dx = dt \hfill \\ \end{gathered}$

$\Rightarrow y = \frac{{{e^{2x}}}}{2} - 3\int {\frac{{ - dt}}{{1 + {{\left( {\frac{{3t}}{{\sqrt 2 }}} \right)}^2}}}}$

$\Rightarrow y = \frac{{{e^{2x}}}}{2} - \sqrt 2 {\tan ^{ - 1}}\frac{{3t}}{{\sqrt 2 }} + c$

$\Rightarrow y = \frac{{{e^{2x}}}}{2} - \sqrt 2 {\tan ^{ - 1}}\frac{{3{e^{ - x}}}}{{\sqrt 2 }} + c$

Step 3 of 3:

If $\left( {0,\frac{1}{2} + \frac{\pi }{{2\sqrt 2 }}} \right)$ satisfies the curve then, we get

$\begin{gathered} \Rightarrow \frac{1}{2} + \frac{\pi }{{2\sqrt 2 }} = \frac{1}{2} + \sqrt 2 {\tan ^{ - 1}}\left( {\frac{3}{{\sqrt 2 }}} \right) + c \hfill \\ \Rightarrow c = \frac{\pi }{{2\sqrt 2 }} - \sqrt 2 {\tan ^{ - 1}}\left( {\frac{3}{{\sqrt 2 }}} \right) \hfill \\ \end{gathered}$

Also, If $\left( {\alpha ,\frac{1}{2}{e^{2\alpha }}} \right)$ satisfies the curve then, we get
$\begin{gathered} \Rightarrow \frac{1}{2}{e^{2\alpha }} = \frac{{{e^{2\alpha }}}}{2} + \sqrt 2 {\tan ^{ - 1}}\left( {\frac{{3{e^{ - \alpha }}}}{{\sqrt 2 }}} \right) + c \hfill \\ \Rightarrow \frac{1}{2}{e^{2\alpha }} = \frac{{{e^{2\alpha }}}}{2} + \sqrt 2 {\tan ^{ - 1}}\left( {\frac{{3{e^{ - \alpha }}}}{{\sqrt 2 }}} \right) + \frac{\pi }{{2\sqrt 2 }} - \sqrt 2 {\tan ^{ - 1}}\left( {\frac{3}{{\sqrt 2 }}} \right) \hfill \\ \Rightarrow {\tan ^{ - 1}}\left( {\frac{3}{{\sqrt 2 }}} \right) - {\tan ^{ - 1}}\left( {\frac{{3{e^{ - \alpha }}}}{{\sqrt 2 }}} \right) = \frac{\pi }{{2\sqrt 2 }} \times \frac{1}{{\sqrt 2 }} \hfill \\ \end{gathered}$

$\Rightarrow {\tan ^{ - 1}}\left( {\frac{{\frac{3}{{\sqrt 2 }} - \frac{{3{e^{ - \alpha }}}}{{\sqrt 2 }}}}{{1 + \frac{9}{2}{e^{ - \alpha }}}}} \right) = \frac{\pi }{4}$ $\left\{ {\because {{\tan }^{ - 1}}(a - b) = {{\tan }^{ - 1}}\left( {\frac{{a - b}}{{1 + ab}}} \right)} \right\}$

$\begin{gathered} \Rightarrow \frac{{\frac{3}{{\sqrt 2 }} - \frac{{3{e^{ - \alpha }}}}{{\sqrt 2 }}}}{{1 + \frac{9}{2}{e^{ - \alpha }}}} = 1 \hfill \\ \Rightarrow \frac{3}{{\sqrt 2 }} - \frac{{3{e^{ - \alpha }}}}{{\sqrt 2 }} = 1 + \frac{9}{2}{e^{ - \alpha }} \hfill \\ \Rightarrow \frac{3}{{\sqrt 2 }}{e^\alpha } - \frac{3}{{\sqrt 2 }} = {e^\alpha } + \frac{9}{2} \hfill \\ \Rightarrow {e^\alpha } = \frac{{\frac{9}{2} + \frac{3}{{\sqrt 2 }}}}{{\frac{3}{{\sqrt 2 }} - 1}} \hfill \\ \Rightarrow {e^\alpha } = \frac{3}{{\sqrt 2 }}\left( {\frac{{3 + \sqrt 2 }}{{3 - \sqrt 2 }}} \right) \hfill \\ \end{gathered}$

Final Answer:

Hence, Option (B) is correct.

Solution:

Step 1 of 3:

Step 2 of 3:

Step 3 of 3:

Final Answer:

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